Question

Angle A is in standard position and terminates in quadrant IV. If sec(A)=4/3, complete the steps to find cot(A).

Answer 1

Answer:

First answer is A.

Second answer is: tan.

Third is answer  C. sqrt7/3

Fourth answer is A. -3sqrt7/7

Step-by-step explanation:

Answer 2

Answer:

Part A) Use the identity $tan^2(A)+1=sec^2(A)$

Part B) $cot(A)=-\frac{3\sqrt{7}}{7}$

Step-by-step explanation:

we know that

$tan^2(A)+1=sec^2(A)$ ----> by trigonometric identity

we have

$sec(A)=\frac{4}{3}$

substitute in the expression above

$tan^2(A)+1=(\frac{4}{3})^2$

solve for tan(A)

$tan^2(A)+1=\frac{16}{9}$

$tan^2(A)=\frac{16}{9}-1$

$tan^2(A)=\frac{7}{9}$

square root both sides

$tan(A)=\pm\frac{\sqrt{7}}{3}$

Remember that Angle A terminates in quadrant IV

so

The value of tan(A) is negative

$tan(A)=-\frac{\sqrt{7}}{3}$

Find the value of cot(A)

we know that

$cot(A)=\frac{1}{tan(A)}$ ----> is the reciprocal

therefore

$cot(A)=-\frac{3}{\sqrt{7}}$

simplify

$cot(A)=-\frac{3\sqrt{7}}{7}$